Free Particle and Wave Packets

Set the potential to zero and quantum mechanics should hand you its easiest problem. Instead it hands you its most instructive paradox. The Schrödinger equation solves instantly — plane waves $e^{i(kx-\omega t)}$ , one for every wavenumber — and yet not a single one of those solutions can be a particle: each is spread uniformly over the entire universe, and Free Particle and Plane Waves shows their normalization integral diverges. The free particle is the system whose stationary states are all unusable. The rescue is superposition, and this article is about watching it work. Add plane waves with neighbouring wavenumbers and something remarkable happens: they agree near one point and argue everywhere else, and a localized lump — a wave packet — condenses out of waves that individually are everywhere. We build the effect by hand, one wave at a time, with a comb of $N$ equally spaced modes: five waves already make a recognizable particle, and the construction is transparent enough to expose everything — why the lump's sharpness is bought with momentum spread, why the whole pattern rides at the group velocity, and why, for a discrete comb, the packet even reassembles itself at a precise revival time. This is the concept-register story of the free particle. The heavy machinery lives next door and is quoted, not rebuilt: the continuum synthesis integral in Wave Packets and Their Construction, the velocity theorems in Group Velocity versus Phase Velocity, the exact spreading law in Wave Packet Spreading and Dispersion. What you take from here is the picture they all rest on — a particle is a chord, not a note.

Key takeaways

The free particle's stationary states are plane waves $e^{i(kx-\omega t)}$ with $\omega = \hbar k^2/2m$ — but none is normalizable, so none is a physical state on its own.
Physical free particles are wave packets: superpositions over a band of wavenumbers, localized where the waves agree, cancelled where they argue.
Each mode's amplitude never changes — it only rotates in phase at its own rate $\omega(k)$ . All packet motion and shape change is choreography of phases.
A comb of $N$ equally spaced modes makes lumps of width $2\pi/N\Delta k$ repeating every $2\pi/\Delta k$ : more waves, sharper particle.
The lump pattern travels at the group velocity $v_g = \hbar k_0/m = p_0/m$ — the classical velocity — while individual crests trail at half that speed.
Because the phase rates $\omega(k)$ are quadratic in $k$ , an equally spaced comb rephases perfectly at $T_{\text{rev}} = 4\pi m/\hbar\,\Delta k^2$ : the packet revives, displaced by exactly $v_g T_{\text{rev}}$ .
Key idea: a free particle is described not by a single plane wave but by a wave packet — a superposition whose envelope moves at the group velocity $p_0/m$ and whose shape evolves because the dispersion relation $\omega = \hbar k^2/2m$ is nonlinear.

The core insight

A free particle is described not by a single plane wave but by a wave packet — a Fourier superposition whose envelope moves at the group velocity $p_0/m$ and inevitably spreads because the dispersion relation $\omega=\hbar k^2/2m$ is nonlinear.

Where this lives today

The comb of this article is not a toy — it is the working principle of the mode-locked laser. Lock $N$ equally spaced optical modes into fixed relative phase and their interference is a train of sharp pulses spaced by the round-trip time: the optical frequency comb, the technology behind attosecond pulse generation and the most precise clocks ever built. The algebra of Derivation B is the algebra of those instruments, with photons in place of electrons. The revival, too, is measured physics. Prepare a Rydberg electron as a superposition of neighbouring energy levels — a discrete comb in energy rather than wavenumber — and its localized orbit packet dephases, blurs, and then snaps back at the revival time, exactly as the quadratic-phase bookkeeping predicts. Pump-probe experiments have filmed these wave-packet revivals since the early 1990s, in atoms and molecules alike; watching a "classical" electron dissolve and reassemble remains one of the cleanest demonstrations that particles are chords of stationary states.

The physical picture

Think of each plane wave as a pure colour: one wavelength, perfectly uniform, everywhere at once. A single colour cannot paint a dot. But overlay several colours whose wavelengths are close: near one special point their crests all line up and the amplitudes add; a little way off, their small wavelength differences have accumulated into disagreement, and they cancel. Localization is not a property any wave has — it is a conspiracy between waves.

Figure

One wave is everywhere — nine waves are somewhere

Two features of the picture repay a careful look. First, the lumps repeat: a finite comb of equally spaced wavenumbers builds a periodic train of particles, spaced $2\pi/\Delta k$ apart. Only in the continuum limit — an unbroken band of wavenumbers, the synthesis integral built in Wave Packets and Their Construction — do the copies retreat to infinity, leaving the single packet of a real experiment. Second, the sharpening is bought, never free: each new mode adds momentum content, so a narrower lump always means a wider spread of momenta. That reciprocity, dressed in $\hbar$ , is the uncertainty principle — stated with its machinery in Standard Deviation and Quantum Uncertainty, made exact in Momentum-Space Wavefunction. And nothing here is static. Each mode's clock runs at its own rate $\omega(k)$ , so the interference pattern is choreography in time: the agreement point drifts (the packet moves), and because the clock rates are not evenly spaced in frequency, the choreography slowly dephases (the packet's shape evolves). Every free-particle phenomenon in this chapter is some aspect of that phase ballet.

From impossible waves to a possible particle

With $V = 0$ , the time-dependent Schrödinger equation of Time-Dependent Schrödinger Equation accepts the travelling wave $e^{i(kx-\omega t)}$ for any $k$ — provided frequency and wavenumber honour one contract. That contract is the system's dispersion relation, and it is nothing but $E = p^2/2m$ wearing wave clothing:

\omega(k) = \frac{\hbar k^2}{2m} \qquad (1)

where: $\omega$ — the phase-rotation rate of the mode $k$ ; the quadratic shape is the free particle's signature (for light, $\omega = ck$ is linear — that difference drives everything below). Via $E = \hbar\omega$ and $p = \hbar k$ from de Broglie Matter Waves, equation (1) is the kinetic-energy relation, mode by mode. No single mode is a state — the flat-modulus verdict and its delta/box normalization repair belong to Free Particle and Plane Waves — but the equation is linear, so any sum of modes solves it too. The general free solution is a superposition over the whole $k$ -line,

\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k)\,e^{i(kx - \hbar k^2 t/2m)}\,\mathrm{d}k \qquad (2)

where: $\phi(k)$ — the recipe saying how much of each mode to pour in, fixed once by the initial shape (the extraction of $\phi$ from $\Psi(x,0)$ , and the craft of choosing it, is the quantitative business of Wave Packets and Their Construction; its momentum meaning is Momentum-Space Wavefunction's, via $\phi(k) = \sqrt{\hbar}\,\Phi(\hbar k)$ ). What this article needs from (2) is one structural fact, proved next: the recipe never changes — only the phases do.

Derivation AEvery pure wave just spins — and superpositions therefore dance

We are establishing the two facts the whole free-particle story rests on: that a single plane wave solves the free Schrödinger equation only when $\omega = \hbar k^2/2m$ , and that in any superposition each amplitude $\phi(k)$ stays fixed forever while its phase rotates at that rate. The strategy is direct substitution — differentiate, cancel the common exponential, read off the contract — followed by a two-mode exhibit showing why fixed amplitudes still produce motion. The thing to watch: the time derivative brings down $-i\omega$ while two space derivatives bring down $-k^2$ , and the $i$ 's conspire so both sides are real multiples of the same mode.

Step 1. Insert $e^{i(kx-\omega t)}$ into $i\hbar\,\partial_t\Psi = -(\hbar^2/2m)\,\partial_x^2\Psi$ :

i\hbar(-i\omega)\,e^{i(kx-\omega t)} = -\frac{\hbar^2}{2m}(ik)^2\,e^{i(kx-\omega t)}

Why: each derivative acts on an exponential, so it only multiplies — the equation becomes algebra.

Step 2. Cancel the exponential and solve: $\hbar\omega = \hbar^2k^2/2m$ , i.e. $\omega = \hbar k^2/2m$ , and no other frequency works. Sympy confirms this is the unique solution. Why: the Schrödinger equation does not accept arbitrary wave shapes and speeds; it enforces the energy-momentum contract mode by mode.

Step 3. Now take any superposition $\Psi = \sum_n c_n\,e^{ik_nx}$ at $t=0$ . Because the equation is linear and each mode evolves by its own phase, the state at time $t$ is $\sum_n c_n\,e^{i(k_nx - \omega_nt)}$ : the coefficients $c_n$ are untouched, and $|c_n(t)| = |c_n|$ exactly. Why: with no potential there is nothing to transfer amplitude between momenta — a force is precisely what changes momentum content, and there is none.

Step 4. Fixed amplitudes do not mean a frozen picture. For two modes, the density works out to $|\Psi|^2 = 2 + 2\cos[(k_1-k_2)x - (\omega_1-\omega_2)t]$ : a moving corrugation. The interference terms inherit the relative phases, and those tick at the rates $\omega_1 - \omega_2$ . Why: densities are built from cross terms, and cross terms beat. How fast the agreement region travels — and why that speed is the classical particle velocity — is the beat algebra owned by Group and Phase Velocity, next in this chapter. *The dispersion solution is unique, every phase factor has unit modulus, and the two-mode density identity holds exactly.*

Building a particle by hand: the comb

Here is the construction the hero figure was drawing. Take $N$ modes of equal amplitude, centred on $k_0$ and equally spaced by $\Delta k$ — wavenumbers $k_n = k_0 + n\,\Delta k$ with $n = -\tfrac{N-1}{2}, \dots, \tfrac{N-1}{2}$ . This "comb" is the simplest packet that can be summed in closed form, and it already contains the whole phenomenology.

Go deeper · Derivation B · 1 minFive waves make a particle — the comb envelope in closed form

Derivation BFive waves make a particle — the comb envelope in closed form

We are summing the $N$ -mode comb at $t=0$ and extracting its intensity pattern exactly. The strategy is to factor out the central carrier $e^{ik_0x}$ , recognize what remains as a finite geometric series in the common ratio $e^{i\Delta k\,x}$ , and sum it. Watch the two length scales that emerge from one formula: the lump width, set by the total bandwidth $N\Delta k$ , and the repeat distance, set by the mode spacing $\Delta k$ — their ratio $N$ is the payoff of adding waves.

Step 1. Factor the carrier and identify the series:

\Psi(x,0) = \sum_{n} e^{i(k_0 + n\Delta k)x} = e^{ik_0x}\sum_{n=-(N-1)/2}^{(N-1)/2} \left(e^{i\Delta k\,x}\right)^{n}

Why: every mode shares the carrier; the packet structure lives entirely in the relative phases $n\Delta k\,x$ .

Step 2. Sum the geometric series (a symmetric one, so the result is real):

\Psi(x,0) = e^{ik_0x}\;\frac{\sin(N\Delta k\,x/2)}{\sin(\Delta k\,x/2)}

Why: the standard finite geometric sum; the symmetric limits cancel the phase offset, leaving the Dirichlet kernel — the same function that governs diffraction gratings, for the same reason: N equally spaced coherent sources.

Step 3. Read the intensity $|\Psi|^2 = \sin^2(N\Delta k\,x/2)/\sin^2(\Delta k\,x/2)$ . At $x = 0$ every mode agrees: the value is $N^2$ — constructive interference squared. The first zero sits where the $N$ phasors first close into a full circle, at $\Delta k\,x = 2\pi/N$ ; the whole pattern repeats when neighbouring phasors regain agreement, at $\Delta k\,x = 2\pi$ . Why: lump width $\sim 2\pi/N\Delta k$ shrinks with the total bandwidth — sharper particles cost more momenta — while the copies at spacing $2\pi/\Delta k$ are the fingerprint of discreteness. A continuous band of $k$ (the integral (2), built in Wave Packets and Their Construction) pushes the copies off to infinity and leaves one packet. *The Dirichlet identity holds for N = 3, 5, and 20: peak exactly N², first zero at 2π/N, period 2π.*

Set the comb in motion by giving each mode its phase $e^{-i\omega_nt}$ from Derivation A, and the lumps travel: the numerical peak of the $N=5$ comb below moves at $1.158\times10^6\ \mathrm{m/s}$ — precisely $\hbar k_0/m = \langle p\rangle/m$ , the classical velocity of a particle with the comb's average momentum, and the speed that Momentum Operator's Ehrenfest card demands of every quantum average. Meanwhile individual crests inside each lump crawl at half that speed and slip backward through the envelope — the two-speed structure whose full anatomy is Group and Phase Velocity's business, next.

Figure

Carrier and envelope — the lump travels at v_g while crests slip through it

The comb's party trick: an exact revival

For a continuous recipe $\phi(k)$ , the quadratic clock rates slowly and permanently dephase the packet — the spreading whose exact law is owned by Wave Packet Spreading and Dispersion. A discrete comb does something a continuum never can: it comes back.

Go deeper · Derivation C · 1 minThe comb reassembles itself — the revival time from quadratic phases

Derivation CThe comb reassembles itself — the revival time from quadratic phases

We are showing that the equally spaced comb re-forms its exact initial shape at $T_{\text{rev}} = 4\pi m/\hbar\,\Delta k^2$ , displaced by one clean multiple of the group velocity. The strategy is pure phase bookkeeping: expand each mode's accumulated phase $\omega_n t$ in the mode index $n$ , and choose $t$ so the piece quadratic in $n$ — the only shape-changing piece — becomes an invisible multiple of $2\pi$ . The linear-in- $n$ piece is harmless: it is exactly a translation. This is why equal spacing matters: only then are the quadratic phases synchronized multiples of one another.

Step 1. Write the phase of mode $n$ at time $t$ , with $k_n = k_0 + n\Delta k$ :

\omega_n t = \frac{\hbar t}{2m}\left(k_0^2 + 2k_0\,n\Delta k + n^2\Delta k^2\right)

Why: the three pieces act differently — a global phase (ignorable), a phase linear in n (a rigid shift of the interference pattern), and a phase quadratic in n (genuine reshaping).

Step 2. Demand the quadratic piece be a multiple of $2\pi$ for every integer $n$ : since $n^2$ is an integer, it suffices that $\frac{\hbar t}{2m}\Delta k^2 = 2\pi$ , giving

T_{\text{rev}} = \frac{4\pi m}{\hbar\,\Delta k^2}

Why: at this instant every shape-changing phase has wound an exact whole number of turns — the modes are back in their original relative configuration, as if no dephasing had happened.

Step 3. Read the surviving linear piece as a displacement: at $T_{\text{rev}}$ the pattern is $|\Psi(x, T_{\text{rev}})|^2 = |\Psi(x - v_g T_{\text{rev}}, 0)|^2$ with $v_g = \hbar k_0/m$ — the reborn packet sits a distance $v_g T_{\text{rev}} = 4\pi k_0/\Delta k^2$ downstream, which is exactly $2k_0/\Delta k$ lump spacings: an integer whenever $2k_0/\Delta k$ is one, so the revived train aligns lump-for-lump with the original lattice. Why: linear-in-n phases factor as $e^{in\Delta k(x - v_gt)}$ — a translation of the envelope, nothing more. All the "damage" of dispersion was stored in the quadratic phases, and they just paid out at par. *For the N = 5, Δk = 10⁹ m⁻¹ electron comb, the density at $T_{\text{rev}}$ matches the initial density to 1 part in 10⁹, while at $T_{\text{rev}}/2$ it differs by more than 30%: the revival is exact and special.*

Limiting cases

case	result	meaning
$N = 1$	flat $\vert\Psi\vert^2$ , no lump	one mode cannot localize; the non-normalizable plane wave of Free Particle and Plane Waves
$N \to \infty$ , band $N\Delta k$ fixed	single packet, copies pushed to infinity	the continuum synthesis integral of Wave Packets and Their Construction takes over
$\Delta k \to 0$ at fixed $N$	lumps widen and separate without bound	vanishing bandwidth means vanishing localization — momentum certainty costs position
linear dispersion $\omega \propto k$ (light in vacuum)	no dephasing ever, no revival needed	all clocks tick proportionally; packets hold shape — dispersion is a quadratic disease
macroscopic mass	shape change invisibly slow	for 1 g localized to 1 mm the dephasing scale $m\sigma^2/\hbar \sim 10^{25}\ \mathrm{s}$ ; the classical particle is a quantum packet that never noticeably dances (exact law: Wave Packet Spreading and Dispersion)

Common trapA wave packet has one definite energy

It does not — and the comb makes the mistake impossible to sustain. The five modes of the worked example carry kinetic energies from 2.44 to 5.49 eV; the packet contains all of them at once, with mean $\langle E\rangle = (\hbar^2/2m)(k_0^2 + \langle n^2\rangle\Delta k^2)$ — strictly larger than $\hbar^2k_0^2/2m$ , because momentum spread is extra kinetic energy. A packet is not "a particle with energy $E$ moving along": it is a coherent chord of energies, and only a single plane wave — which is not a physical state — plays one note. When an exam asks "the energy of the packet," it is asking for $\langle E\rangle$ or the band, never a single sharp value; sharp energy belongs to stationary states, and the free particle has none worth the name.

Worked examples

Example 1 — build the comb, read every scale off it (confidence builder). An electron packet is synthesized from $N = 5$ equal-amplitude modes centred at $k_0 = 1.00\times10^{10}\ \mathrm{m^{-1}}$ , spaced $\Delta k = 1.00\times10^{9}\ \mathrm{m^{-1}}$ . Find the lump spacing, the central-lobe half-width, the carrier wavelength, both speeds, and the energy content.

Step 1. Geometry from the Dirichlet formula: lumps repeat every $2\pi/\Delta k = \mathbf{6.28\ nm}$ ; the central lobe's first zero sits at $2\pi/N\Delta k = \mathbf{1.26\ nm}$ from the peak; the carrier ripples inside have wavelength $2\pi/k_0 = \mathbf{0.628\ nm}$ — ten crests per repeat distance.

Step 2. Speeds: the envelope travels at $v_g = \hbar k_0/m_e = \mathbf{1.16\times10^{6}\ m/s}$ (numerically tracked: the peak moves at exactly this rate), the crests at $v_p = \hbar k_0/2m_e = \mathbf{5.79\times10^{5}\ m/s}$ — half. Why the particle is the envelope, and what the factor of two means, is Group and Phase Velocity's story.

Step 3. Energy content: the five modes carry $E_n = \hbar^2k_n^2/2m_e$ = 2.44, 3.09, 3.81, 4.61, 5.49 eV — a 3.05 eV band around the carrier's 3.81 eV, with mean $\langle E\rangle = \mathbf{3.886\ eV}$ . The packet's mean energy exceeds the carrier energy by the spread term: localization costs kinetic energy.

Example 2 — the revival, timed and placed (exam level). For the same comb, find the revival time, how far the packet has travelled when it revives, and verify the revived train lands on the original lattice.

Step 1. Revival time from Derivation C: $T_{\text{rev}} = 4\pi m_e/\hbar\Delta k^2 = 4\pi(9.109\times10^{-31})/(1.055\times10^{-34})(10^{9})^2 = \mathbf{108.5\ fs}$ .

Step 2. Displacement: $v_g T_{\text{rev}} = (1.158\times10^{6})(1.085\times10^{-13}) = \mathbf{125.7\ nm}$ .

Step 3. Lattice check: $125.7\ \mathrm{nm} / 6.28\ \mathrm{nm} = 2k_0/\Delta k = \mathbf{20.0}$ lump spacings exactly — an integer, so the revived pattern coincides with the original one lump-for-lump. Numerically the two densities agree to one part in $10^9$ ; at $t = T_{\text{rev}}/2$ they disagree by over 30%. The revival is exact, and exactly scheduled.

Origins · 1 minPeter Debye and how this was discovered

Peter Debye

1884–1966 · Zurich, then Berlin and Cornell

The wave packet has a godfather, and it is the man who demanded the wave equation exist. In late 1925, Debye — then Zurich's imperious professor of physics — assigned his colloquium the task of reporting on de Broglie's strange thesis. Schrödinger gave the talk; Debye's verdict, as Felix Bloch later recalled it, was that the whole business was childish — "to deal properly with waves, one must have a wave equation." Weeks later Schrödinger returned with one, and the free packet of this article became writable. The prompt was in character: Debye had spent 1909 building the asymptotic saddle-point machinery that tracks where superposed waves constructively interfere — essentially the mathematics of packet peaks — for Sommerfeld's optics problems. He collected his own Nobel in 1936, in chemistry, for dipole moments and X-ray scattering, having drifted from the quantum storm he helped trigger; the packet, the colloquium, and the missing equation remain his accidental legacy to every chapter like this one.

History

1909

— Debye's saddle-point asymptotics: the mathematics of where superposed waves agree.

1924

— de Broglie attaches a wave to every particle; no equation yet governs it.

1925

— the Zurich colloquium: Debye tells Schrödinger that waves need a wave equation.

1926

— Schrödinger publishes it, and builds the first coherent packet for the oscillator.

1929

— Darwin works out free-packet motion and spreading in detail; the dispersing electron becomes textbook material.

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.4 (The Free Particle) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Ch. I & Complement G_I · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §1.6–1.7 · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6 The dispersion relation follows from substituting a plane wave into the free TDSE; the Dirichlet-kernel comb sum and its scales have been checked numerically at N = 3, 5, 20; the speeds $v_g = \hbar k_0/m = p_0/m$ and $v_p = v_g/2$ follow from the dispersion relation; and the comb revival $T_{\text{rev}} = 4\pi m/\hbar\Delta k^2$ follows from the phase bookkeeping of Derivation C, numerically exact to $10^{-9}$ . The continuum synthesis, Gaussian packet craft, stationary-phase theorem, and spreading law are treated in their respective articles and quoted here. $\hbar = 1.054571817\times10^{-34}\ \mathrm{J\,s}$ (CODATA).

Builds on: Time-Dependent Schrödinger Equation · Time-Independent Schrödinger Equation · The Wavefunction · de Broglie Matter Waves · Born Statistical Interpretation · Probability Density and Normalization

Leads to: Stationary States and Separation of Variables · Double-Slit Experiment · Wave–Particle Duality

Real packets are continuous superpositions — why does this article use a discrete comb?

Because the comb is the packet you can hold in your hand: the sum closes in three lines, every scale is visible in the formula, and the dynamics (drift, dephasing, revival) is pure phase bookkeeping. The continuum is the $N\to\infty$ limit — copies pushed to infinity, revival postponed forever — and its machinery (the synthesis integral, Gaussian craft) is built properly in Wave Packets and Their Construction. Concepts here, calculus there.

Is the free particle's energy spectrum discrete or continuous?

Continuous: every $E = \hbar^2k^2/2m \geq 0$ is allowed, which is precisely why no normalizable eigenstate exists (a discrete bound state needs a confining potential to select it). The comb chooses a discrete subset of modes for pedagogy; nature does not. The bound/scattering dichotomy behind this is drawn in Bound States versus Scattering States.

If each mode's amplitude never changes, in what sense does the packet spread?

Spreading is entirely a phase phenomenon. The momentum distribution $|\phi(k)|^2$ is rigorously static — nothing about the particle's momentum content evolves. But the position pattern is an interference effect among the modes, and interference cares about relative phases, which drift quadratically. The frozen-momentum, spreading-position picture is worked exactly in Wave Packet Spreading and Dispersion.

Does anything physical travel at the phase velocity?

No signal and no probability does. Individual crests move at $v_p = \hbar k/2m$ , but a crest is not a thing — it is a feature of the interference pattern, and it carries neither energy nor information. The particle — the probability lump, the expectation value, the detectable object — moves at $v_g = p/m$ . The full two-velocity anatomy, including why $v_p$ even depends on where you set the zero of energy, is Group and Phase Velocity's subject.

Will a real free electron in the lab ever revive like the comb?

Not by itself: a genuinely free electron has a continuous $\phi(k)$ , the quadratic phases never resynchronize, and the spreading is permanent. Revivals need discreteness — equally spaced (or nearly so) components. That is exactly what bound systems supply: Rydberg atoms, molecular vibrations, and particles in boxes all show measured wave-packet revivals. The free comb is the cleanest model of that physics, not a prediction for free space.

The packet peak is where the particle is, right?

The peak is where the particle is most likely to be found — nothing more. One detection event returns one point, distributed according to $|\Psi|^2$ ; the packet is a probability amplitude, not a smeared object. The interpretive contract is Born Statistical Interpretation's, and the packet inherits it unchanged: build the state from waves, then read $|\Psi|^2$ as betting odds for a point particle.