---
title: Ladder Operators and the Number States
module: The Oscillator Algebraically, and Symmetry
moduleNumber: 5
lessonNumber: 1
order: 501
summary: >
  The harmonic oscillator can be solved without touching a differential equation.
  Factoring the Hamiltonian into a lowering operator and its adjoint turns the
  spectrum into pure algebra: the commutator relation fixes the ladder, the
  vacuum condition fixes the ground state, and the energies fall out as
  equally spaced rungs. The same operators give the matrix elements of position
  and momentum for free.
topics: [The Oscillator Algebraically, and Symmetry]
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 2 — Time-Independent Schrödinger Equation; §2.3.1 The Algebraic Method"
  - book: Sakurai & Napolitano
    ref: "Ch. 2 — Quantum Dynamics; §2.3 The Simple Harmonic Oscillator"
  - book: Shankar
    ref: "Ch. 7 — The Harmonic Oscillator; §7.4 The Oscillator in the Energy Basis"
draft: false
---

The [analytic solution](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator)
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 $E_n = (n+\tfrac12)\hbar\omega$ 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.[^gs-alg]

## Factoring the Hamiltonian

The one-dimensional oscillator Hamiltonian is

$$
\hat H = \frac{\hat p^{2}}{2m} + \tfrac12 m\omega^{2}\hat x^{2},
\qquad [\hat x,\hat p] = i\hbar .
$$

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

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

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

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

The oscillator Hamiltonian is therefore

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

Reversing the product order and using $[\hat x,\hat p]=i\hbar$ once more gives
$\hat a\hat a^{\dagger} = \hat H/\hbar\omega + \tfrac12$. Subtracting the two
expressions leaves the one relation that drives everything:

> **Theorem (Canonical commutator of the ladder operators).** The lowering and
> raising operators of the harmonic oscillator satisfy
> $$
> [\hat a,\hat a^{\dagger}] = 1 .
> $$
> Every property of the spectrum below is a consequence of this relation together
> with $\hat H = \hbar\omega(\hat a^{\dagger}\hat a + \tfrac12)$.

## The number operator

Define the **number operator** $\hat N = \hat a^{\dagger}\hat a$, so that
$\hat H = \hbar\omega(\hat N + \tfrac12)$. It is Hermitian and positive: for any
state $\lvert\psi\rangle$,
$\langle\psi\rvert\hat N\lvert\psi\rangle = \lVert \hat a\lvert\psi\rangle\rVert^{2}\ge 0$.
Since $\hat H$ and $\hat N$ differ only by a scale and a shift, they share every
eigenstate; label them by the eigenvalue $\nu$,

$$
\hat N\lvert\nu\rangle = \nu\lvert\nu\rangle,
\qquad
\hat H\lvert\nu\rangle = \hbar\omega\left(\nu + \tfrac12\right)\lvert\nu\rangle .
$$

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

$$
[\hat N,\hat a] = [\hat a^{\dagger}\hat a,\hat a] = [\hat a^{\dagger},\hat a]\,\hat a = -\hat a,
\qquad
[\hat N,\hat a^{\dagger}] = +\hat a^{\dagger}.
$$

These two relations are the ladder. Apply $\hat N$ to the state $\hat a^{\dagger}\lvert\nu\rangle$:

$$
\hat N\bigl(\hat a^{\dagger}\lvert\nu\rangle\bigr)
= \bigl(\hat a^{\dagger}\hat N + \hat a^{\dagger}\bigr)\lvert\nu\rangle
= (\nu+1)\,\hat a^{\dagger}\lvert\nu\rangle .
$$

So $\hat a^{\dagger}\lvert\nu\rangle$ 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 $\hat a\lvert\nu\rangle$ carries eigenvalue $\nu-1$.
The operators shift the energy by exactly one quantum $\hbar\omega$ in either
direction.

$$
% caption: The number states form a ladder spaced by $\hbar\omega$. The raising
% operator $\hat a^{\dagger}$ carries $\lvert n\rangle$ up one rung and the
% lowering operator $\hat a$ carries it down; the bottom rung is the vacuum
% $\lvert 0\rangle$, annihilated by $\hat a$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \n/\y in {0/0, 1/1.2, 2/2.4, 3/3.6} {
    \draw[thick] (0,\y) -- (3.4,\y);
    \node[anchor=west] at (3.5,\y) {$n = \n$};
    \node[anchor=east, font=\scriptsize] at (-0.1,\y) {$E_\n$};
  }
  % highlight the vacuum rung on which the ladder terminates
  \draw[acc, thick] (0,0) -- (3.4,0);
  \node[acc, anchor=west] at (3.5,0) {$n = 0$};
  \node[anchor=west, font=\scriptsize] at (3.5,4.8) {$\vdots$};
  \draw[thick, dotted] (0,4.4) -- (3.4,4.4);
  % raising arrows on the right side
  \draw[->, black!70, thick] (2.6,0.15) -- (2.6,1.05);
  \draw[->, black!70, thick] (2.6,1.35) -- (2.6,2.25);
  \draw[->, black!70, thick] (2.6,2.55) -- (2.6,3.45);
  \node[black!70, anchor=west, font=\scriptsize] at (2.7,0.6) {raise};
  % lowering arrows on the left side
  \draw[->, black, thick] (0.7,1.05) -- (0.7,0.15);
  \draw[->, black, thick] (0.7,2.25) -- (0.7,1.35);
  \draw[->, black, thick] (0.7,3.45) -- (0.7,2.55);
  \node[black, anchor=east, font=\scriptsize] at (0.6,3.0) {lower};
\end{tikzpicture}
$$

## 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 $\lvert 0\rangle$ with

$$
\hat a\lvert 0\rangle = 0 .
$$

This state has $\hat N\lvert 0\rangle = \hat a^{\dagger}\hat a\lvert 0\rangle = 0$,
so its eigenvalue is $\nu = 0$ exactly. Suppose instead the smallest surviving
eigenvalue were some $\nu_0$ not equal to zero with $\hat a\lvert\nu_0\rangle\neq0$.
Then $\hat a\lvert\nu_0\rangle$ 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

$$
\nu = n \in \{0,1,2,\dots\},
\qquad
E_n = \left(n+\tfrac12\right)\hbar\omega .
$$

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.

$$
% caption: The lowering operator applied to the vacuum gives zero, terminating the
% descent. Positivity of $\hat N$ forbids negative rungs, so the ladder must end on
% the state $\hat a\lvert 0\rangle = 0$ at eigenvalue $n=0$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[acc, thick] (0,0) -- (3.2,0);
  \node[acc, anchor=west] at (3.3,0) {$n = 0$ (vacuum)};
  \draw[thick] (0,1.2) -- (3.2,1.2);
  \node[anchor=west] at (3.3,1.2) {$n = 1$};
  % forbidden region below the vacuum
  \fill[black] (0,-1.6) rectangle (3.2,-0.2);
  \draw[black, thick, dashed] (0,-1.2) -- (3.2,-1.2);
  \node[black, anchor=west, font=\scriptsize] at (3.3,-1.2) {forbidden ($n < 0$)};
  % lowering arrow hitting zero
  \draw[->, black!70, thick] (1.6,1.05) -- (1.6,0.15);
  \node[black!70, anchor=west, font=\scriptsize] at (1.7,0.6) {annihilate};
  \node[black, anchor=north, font=\scriptsize] at (1.6,-0.05) {lowering terminates};
\end{tikzpicture}
$$

## Normalization and the tower

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

$$
\lVert \hat a\lvert n\rangle\rVert^{2}
= \langle n\rvert \hat a^{\dagger}\hat a\lvert n\rangle = n,
\qquad
\lVert \hat a^{\dagger}\lvert n\rangle\rVert^{2}
= \langle n\rvert \hat a\hat a^{\dagger}\lvert n\rangle = n+1 .
$$

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

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

The factor $\sqrt n$ in the lowering formula reproduces $\hat a\lvert0\rangle = 0$
automatically. Applying the raising operator $n$ times to the vacuum and dividing by
the accumulated norms builds every state:

> **Theorem (Number states from the vacuum).** The normalized energy eigenstates are
> $$
> \lvert n\rangle = \frac{(\hat a^{\dagger})^{n}}{\sqrt{n!}}\,\lvert 0\rangle,
> \qquad n = 0,1,2,\dots
> $$
> They are orthonormal, $\langle m\vert n\rangle = \delta_{mn}$, and complete on the
> oscillator Hilbert space. Each carries energy $E_n = (n+\tfrac12)\hbar\omega$.

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 $\hat N$.

$$
% caption: Repeated action of the creation operator $\hat a^{\dagger}$ generates the
% whole tower of number states $\lvert n\rangle$ from the vacuum; the normalization
% $\lvert n\rangle = (\hat a^{\dagger})^{n}\lvert 0\rangle/\sqrt{n!}$ collects the
% $\sqrt{k}$ factors picked up at each rung.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  st/.style={draw, minimum width=15mm, minimum height=8mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[st, acc, fill=acc!12] (v) at (0,0) {$n = 0$};
  \node[st] (o) at (2.6,0) {$n = 1$};
  \node[st] (t) at (5.2,0) {$n = 2$};
  \node[st] (h) at (7.8,0) {$n = 3$};
  \node[anchor=west] at (9.1,0) {...};
  \draw[->, black!70, thick] (v) -- (o) node[midway, above, font=\scriptsize] {create};
  \draw[->, black!70, thick] (o) -- (t) node[midway, above, font=\scriptsize] {create};
  \draw[->, black!70, thick] (t) -- (h) node[midway, above, font=\scriptsize] {create};
  \node[black, anchor=north, font=\scriptsize] at (0,-0.7) {$E_0$};
  \node[black, anchor=north, font=\scriptsize] at (7.8,-0.7) {$E_3$};
\end{tikzpicture}
$$

## Position and momentum as ladder combinations

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

$$
\hat x = \sqrt{\frac{\hbar}{2m\omega}}\,\bigl(\hat a + \hat a^{\dagger}\bigr),
\qquad
\hat p = i\sqrt{\frac{m\hbar\omega}{2}}\,\bigl(\hat a^{\dagger} - \hat a\bigr).
$$

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

$$
\langle m\rvert\hat x\lvert n\rangle
= \sqrt{\frac{\hbar}{2m\omega}}
  \Bigl(\sqrt{n}\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1}\Bigr).
$$

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](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator):
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 $\hbar\omega$.

$$
% caption: The position operator in the number basis is a tridiagonal band: the
% only nonzero entries $\langle m\rvert\hat x\lvert n\rangle$ sit one step off the
% diagonal, weighted by $\sqrt{n}$ and $\sqrt{n+1}$. Each cell shows the integer $k$
% under the entry's root $\sqrt{k}$; the empty diagonal encodes
% $\langle n\rvert\hat x\lvert n\rangle = 0$.
\begin{tikzpicture}[>=stealth, font=\scriptsize]
  \definecolor{acc}{HTML}{4A6FA5}
  % 4x4 grid
  \foreach \i in {0,1,2,3,4} {
    \draw[black] (0,\i) -- (4,\i);
    \draw[black] (\i,0) -- (\i,4);
  }
  % row/col headers (n increases downward and rightward)
  \foreach \c/\lab in {0/0,1/1,2/2,3/3} \node[black] at (\c+0.5,4.3) {\lab};
  \foreach \r/\lab in {0/3,1/2,2/1,3/0} \node[black] at (-0.3,\r+0.5) {\lab};
  \node at (-0.3,4.3) {$m$};
  \node at (4.3,4.3) {$n$};
  % off-diagonal filled cells: pairs (m=n-1) and (m=n+1)
  % grid row index from bottom: r=0 -> m=3 ... r=3 -> m=0
  % entry (m,n) nonzero if |m-n|=1
  \foreach \m/\n in {0/1, 1/0, 1/2, 2/1, 2/3, 3/2} {
    \fill[acc!22] (\n,3-\m) rectangle (\n+1,4-\m);
    \draw[acc] (\n,3-\m) rectangle (\n+1,4-\m);
  }
  \node at (1.5,3.5) {$1$};
  \node at (0.5,2.5) {$1$};
  \node at (2.5,2.5) {$2$};
  \node at (1.5,1.5) {$2$};
  \node at (3.5,1.5) {$3$};
  \node at (2.5,0.5) {$3$};
\end{tikzpicture}
$$

Each nonzero cell shows the integer $k$ whose root $\sqrt{k}$ is the matrix element;
the common prefactor $\sqrt{\hbar/2m\omega}$ 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 $\hat a\lvert 0\rangle = 0$ is a first-order differential
equation once $\hat a$ is written with $\hat p = -i\hbar\,\d/\d x$:

$$
\sqrt{\frac{m\omega}{2\hbar}}\left(x + \frac{\hbar}{m\omega}\frac{\d}{\d x}\right)\psi_0(x) = 0
\quad\Longrightarrow\quad
\frac{\d\psi_0}{\d x} = -\frac{m\omega}{\hbar}\,x\,\psi_0 .
$$

Separating and integrating,

$$
\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}
\exp\!\left(-\frac{m\omega}{2\hbar}\,x^{2}\right),
$$

with the prefactor fixed by $\int|\psi_0|^{2}\,\d x = 1$. This is the Gaussian
ground state of the
[analytic treatment](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator).
Every excited state then follows by applying the differential form of
$\hat a^{\dagger}$, 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 $H_n$.

> **Worked example.** Compute $\langle x^{2}\rangle$ and $\langle p^{2}\rangle$ in
> the state $\lvert n\rangle$ directly from the ladder algebra.
>
> Write $\hat x^{2} = \dfrac{\hbar}{2m\omega}(\hat a + \hat a^{\dagger})^{2}
> = \dfrac{\hbar}{2m\omega}(\hat a^{2} + \hat a^{\dagger 2} + \hat a\hat a^{\dagger} + \hat a^{\dagger}\hat a)$.
> The terms $\hat a^{2}$ and $\hat a^{\dagger 2}$ shift $\lvert n\rangle$ to
> $\lvert n\pm2\rangle$, orthogonal to $\lvert n\rangle$, so they drop from the
> expectation value. The survivors give
> $$
> \langle n\rvert\hat x^{2}\lvert n\rangle
> = \frac{\hbar}{2m\omega}\,\langle n\rvert(\hat a\hat a^{\dagger} + \hat a^{\dagger}\hat a)\lvert n\rangle
> = \frac{\hbar}{2m\omega}\,(2n+1).
> $$
> The same steps with $\hat p^{2} = -\dfrac{m\hbar\omega}{2}(\hat a^{\dagger}-\hat a)^{2}$
> give $\langle p^{2}\rangle = \tfrac12 m\hbar\omega\,(2n+1)$.

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

$$
\langle T\rangle = \frac{\langle p^{2}\rangle}{2m} = \frac12\left(n+\tfrac12\right)\hbar\omega
= \langle V\rangle = \tfrac12 m\omega^{2}\langle x^{2}\rangle,
$$

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

$$
\Delta x\,\Delta p = \sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}
= \left(n+\tfrac12\right)\hbar \ \ge\ \frac{\hbar}{2},
$$

saturated at $n=0$. The Gaussian vacuum is the minimum-uncertainty state, and the
$\tfrac12\hbar\omega$ ground-state energy is precisely the energy of that
minimum-uncertainty compromise between kinetic and potential terms.

$$
% caption: Kinetic and potential energy share $E_n$ equally at every level (quantum
% virial theorem), while the uncertainty product $\Delta x\,\Delta p = (n+\tfrac12)\hbar$
% grows one $\hbar$ per rung, minimized by the vacuum.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (0,0) -- (5.6,0) node[right, font=\scriptsize, text=black] {$n$};
  \draw[black] (0,0) -- (0,3.4) node[above, font=\scriptsize, text=black] {energy};
  \foreach \n in {0,1,2,3} {
    \node[black, anchor=north, font=\scriptsize] at (\n*1.4+0.7,0) {\n};
    % total energy bar
    \draw[acc, very thick] (\n*1.4+0.35,{(\n+0.5)*0.8}) -- (\n*1.4+1.05,{(\n+0.5)*0.8});
    % half-marks for T and V
    \fill[black] (\n*1.4+0.7,{(\n+0.5)*0.4}) circle (2pt);
  }
  \node[acc, anchor=west, font=\scriptsize] at (4.3,3.0) {$E_n$};
  \node[black, anchor=west, font=\scriptsize] at (1.3,3.05) {$T = V = \tfrac12 E_n$};
\end{tikzpicture}
$$

## The ladder operators in the Heisenberg picture

The same commutator that fixed the spectrum also fixes the dynamics. In the
[Heisenberg picture](/quantum-mechanics/formalism/time-evolution-schrodinger-and-heisenberg-pictures)
an operator evolves by $\dot{\hat A} = (i/\hbar)[\hat H,\hat A]$. For the lowering
operator, using $\hat H = \hbar\omega(\hat a^\dagger\hat a + \tfrac12)$ and
$[\hat a^\dagger\hat a,\hat a] = -\hat a$,

$$
\dot{\hat a} = \frac{i}{\hbar}[\hat H,\hat a]
= i\omega\,[\hat a^\dagger\hat a,\hat a]
= -i\omega\,\hat a
\quad\Longrightarrow\quad
\hat a(t) = \hat a(0)\,e^{-i\omega t},
$$

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

$$
\hat x(t) = \hat x(0)\cos\omega t + \frac{\hat p(0)}{m\omega}\sin\omega t,
\qquad
\hat p(t) = \hat p(0)\cos\omega t - m\omega\,\hat x(0)\sin\omega t .
$$

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 $\lvert n\rangle$ has $\langle\hat x\rangle = 0$
at all times because $\hat a$ and $\hat a^\dagger$ connect it only to orthogonal
neighbors. Producing a moving $\langle\hat x(t)\rangle$ 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](/quantum-mechanics/oscillator-and-symmetry/coherent-and-squeezed-states).

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

| Feature | Analytic (Hermite) method | Algebraic (ladder) method |
| --- | --- | --- |
| Primary object | eigenfunctions $\psi_n(x)$ | eigenstates $\lvert n\rangle$ |
| Spectrum from | series-termination condition | commutator + positivity of $\hat N$ |
| Ground state from | Gaussian asymptotics | $\hat a\lvert 0\rangle = 0$ |
| Excited states from | recursion for $H_n$ | $(\hat a^\dagger)^n\lvert 0\rangle/\sqrt{n!}$ |
| Matrix elements | integrals of $H_n\,x\,H_m$ | read off $\hat a,\hat a^\dagger$ |
| Generalizes to | one-dimensional potentials | any bosonic mode / field |

## Why the algebra generalizes

The construction used only three inputs: a Hamiltonian written as $\hbar\omega(\hat N+\tfrac12)$
with $\hat N=\hat a^\dagger\hat a$, the commutator $[\hat a,\hat a^{\dagger}]=1$, 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 $\hat a^{\dagger}$, $\hat a$ become the **creation** and
**annihilation** operators that add or remove one quantum of excitation. The number
state $\lvert n\rangle$ is then read as "$n$ quanta present," and the vacuum
$\lvert 0\rangle$ as the state with none. The
[coherent states](/quantum-mechanics/oscillator-and-symmetry/coherent-and-squeezed-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.

[^gs-alg]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ (3rd ed., Cambridge, 2018), §2.3.1 — the algebraic method: the ladder operators, the commutator $[\hat a,\hat a^{\dagger}]=1$, the vacuum condition $\hat a\lvert0\rangle=0$, and the spectrum $E_n=(n+\tfrac12)\hbar\omega$ obtained without solving the differential equation. Cambridge listing: https://www.cambridge.org/highereducation/books/introduction-to-quantum-mechanics/990799CA07A83FC5312402AF6860311E. Parallel treatments: **Sakurai & Napolitano**, _Modern Quantum Mechanics_ (3rd ed., Cambridge, 2021), §2.3; **Shankar**, _Principles of Quantum Mechanics_ (2nd ed., Springer, 1994), §7.4 — https://link.springer.com/book/10.1007/978-1-4757-0576-8.